Grade 9 Mathematics (MTH1W)
MTH1W is Ontario's de-streamed Grade 9 Mathematics course, meaning all students — regardless of future academic plans — study the same rich curriculum together. The course builds algebraic reasoning, geometric thinking, financial literacy, and data skills that underpin every subsequent mathematics course in the Ontario secondary curriculum.
1. Number Sense: Powers and Rational Numbers
A solid understanding of number is essential before moving into algebra. In Grade 9 you extend your work with integers, fractions, and decimals to include rational numbers and powers with integer exponents.
Rational Numbers
Definition: Rational Number
A rational number is any number that can be written in the form $\dfrac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. The set of rational numbers is denoted $\mathbb{Q}$.
Examples: $\dfrac{3}{4}$, $-\dfrac{7}{2}$, $0.6$ (since $0.6 = \dfrac{3}{5}$), $-3$ (since $-3 = \dfrac{-3}{1}$).
Operations with rational numbers follow the same rules as fractions, extended to negative values. When adding fractions, always find a common denominator first.
Worked Example 1.1 — Operations with Rational Numbers
Evaluate $\dfrac{3}{4} - \dfrac{-5}{6} + \dfrac{1}{3}$.
Step 1 Find the least common denominator (LCD) of 4, 6, and 3. The LCD is 12.
Step 2 Convert each fraction: $$\frac{3}{4} = \frac{9}{12}, \quad \frac{-5}{6} = \frac{-10}{12}, \quad \frac{1}{3} = \frac{4}{12}$$
Step 3 Subtracting a negative is the same as adding: $$\frac{9}{12} - \frac{-10}{12} + \frac{4}{12} = \frac{9 + 10 + 4}{12} = \frac{23}{12}$$
Powers with Integer Exponents
Exponent Laws
For any non-zero base $a$ and integers $m$, $n$:
- Product rule: $a^m \cdot a^n = a^{m+n}$
- Quotient rule: $a^m \div a^n = a^{m-n}$
- Power rule: $(a^m)^n = a^{mn}$
- Zero exponent: $a^0 = 1$
- Negative exponent: $a^{-n} = \dfrac{1}{a^n}$
Worked Example 1.2 — Simplifying with Exponent Laws
Simplify $\dfrac{(2x^3)^2 \cdot x^{-4}}{4x}$.
Step 1 Apply the power rule to the numerator: $(2x^3)^2 = 2^2 \cdot x^6 = 4x^6$.
Step 2 Combine in the numerator using the product rule: $4x^6 \cdot x^{-4} = 4x^{6+(-4)} = 4x^2$.
Step 3 Divide: $\dfrac{4x^2}{4x} = x^{2-1} = x$.
The simplified answer is $x$.
Scientific Notation
Very large or very small numbers are expressed in scientific notation as $a \times 10^n$ where $1 \leq a < 10$ and $n$ is an integer. For example, the distance from Earth to the Sun is approximately $1.496 \times 10^{11}$ metres.
2. Algebra: Linear Equations and Polynomials
Solving Linear Equations
A linear equation in one variable has the form $ax + b = c$. To solve, isolate the variable by performing inverse operations in reverse order of BEDMAS (apply addition/subtraction first, then multiplication/division).
Worked Example 2.1 — Solving a Linear Equation
Solve $3(2x - 4) = 2x + 8$.
Step 1 Expand the left side: $6x - 12 = 2x + 8$.
Step 2 Move variable terms to the left: $6x - 2x = 8 + 12$, giving $4x = 20$.
Step 3 Divide both sides by 4: $x = 5$.
Check $3(2(5) - 4) = 3(6) = 18$ and $2(5) + 8 = 18$. ✓
Solving Linear Inequalities
Linear inequalities are solved exactly like equations, with one important rule: when you multiply or divide both sides by a negative number, you must reverse the inequality sign.
Worked Example 2.2 — Linear Inequality
Solve $-2x + 5 > 11$ and graph the solution on a number line.
Step 1 Subtract 5 from both sides: $-2x > 6$.
Step 2 Divide by $-2$ and reverse the inequality: $x < -3$.
The solution is all real numbers less than $-3$, written in interval notation as $(-\infty, -3)$. On a number line, draw an open circle at $-3$ and shade to the left.
Polynomials
Polynomial Vocabulary
A polynomial is an algebraic expression consisting of terms of the form $ax^n$ where $a$ is a real number (the coefficient) and $n$ is a non-negative integer (the degree of that term).
- Monomial: one term — e.g., $5x^2$
- Binomial: two terms — e.g., $3x - 7$
- Trinomial: three terms — e.g., $x^2 + 2x - 8$
The degree of a polynomial is the highest degree among all its terms.
Adding and Subtracting Polynomials
Collect like terms — terms with the same variable raised to the same power. Be careful with signs when subtracting.
Worked Example 2.3 — Polynomial Operations
Simplify $(3x^2 - 5x + 2) - (x^2 + 3x - 7)$.
Step 1 Distribute the negative sign: $3x^2 - 5x + 2 - x^2 - 3x + 7$.
Step 2 Collect like terms: $$(3x^2 - x^2) + (-5x - 3x) + (2 + 7) = 2x^2 - 8x + 9$$
Multiplying Polynomials
Use the distributive property (expand and simplify). When multiplying two binomials, you can use FOIL (First, Outer, Inner, Last).
Worked Example 2.4 — Expanding a Product
Expand and simplify $(2x + 3)(x - 5)$.
FOIL $$\underbrace{(2x)(x)}_{\text{First}} + \underbrace{(2x)(-5)}_{\text{Outer}} + \underbrace{(3)(x)}_{\text{Inner}} + \underbrace{(3)(-5)}_{\text{Last}}$$ $$= 2x^2 - 10x + 3x - 15 = 2x^2 - 7x - 15$$
3. Analytic Geometry: Linear Relations
Analytic geometry connects algebra and geometry by placing geometric objects on a coordinate plane. The key idea is that a linear equation in two variables describes a straight line.
Slope of a Line
Definition: Slope
The slope $m$ of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is: $$m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}, \quad x_1 \neq x_2$$
- Positive slope: line rises from left to right
- Negative slope: line falls from left to right
- Zero slope: horizontal line
- Undefined slope: vertical line
Worked Example 3.1 — Finding Slope
Find the slope of the line passing through $(-1, 4)$ and $(3, -2)$.
Solution $$m = \frac{-2 - 4}{3 - (-1)} = \frac{-6}{4} = -\frac{3}{2}$$
The slope is $-\dfrac{3}{2}$, meaning for every 2 units moved right, the line drops 3 units.
Forms of a Linear Equation
Key Forms
- Slope-intercept form: $y = mx + b$ ($m$ = slope, $b$ = $y$-intercept)
- Standard form: $Ax + By + C = 0$ (where $A$, $B$, $C$ are integers)
- Point-slope form: $y - y_1 = m(x - x_1)$ (useful when you know a point and the slope)
Worked Example 3.2 — Writing the Equation of a Line
Write the equation of the line with slope $\dfrac{2}{3}$ passing through the point $(3, -1)$ in slope-intercept form.
Step 1 Use point-slope form: $y - (-1) = \dfrac{2}{3}(x - 3)$.
Step 2 Expand: $y + 1 = \dfrac{2}{3}x - 2$.
Step 3 Solve for $y$: $y = \dfrac{2}{3}x - 3$.
Parallel and Perpendicular Lines
Two lines are parallel if they have the same slope ($m_1 = m_2$) and different $y$-intercepts. Two lines are perpendicular if their slopes are negative reciprocals: $m_1 \cdot m_2 = -1$, equivalently $m_2 = -\dfrac{1}{m_1}$.
Worked Example 3.3 — Perpendicular Lines
Find the equation of the line perpendicular to $y = \dfrac{3}{4}x - 2$ that passes through $(6, 1)$.
Step 1 The given slope is $m_1 = \dfrac{3}{4}$. The perpendicular slope is $m_2 = -\dfrac{4}{3}$.
Step 2 Use point-slope: $y - 1 = -\dfrac{4}{3}(x - 6)$.
Step 3 Expand and simplify: $y = -\dfrac{4}{3}x + 8 + 1 = -\dfrac{4}{3}x + 9$.
4. Measurement and Geometry
Perimeter, Area, and Volume
Grade 9 extends measurement to three-dimensional objects. The key formulas you need to know:
Key Formulas
- Circle: Circumference $= 2\pi r$; Area $= \pi r^2$
- Rectangle: Perimeter $= 2(l + w)$; Area $= lw$
- Triangle: Area $= \dfrac{1}{2}bh$
- Rectangular prism: Surface area $= 2(lw + lh + wh)$; Volume $= lwh$
- Cylinder: Surface area $= 2\pi r^2 + 2\pi rh$; Volume $= \pi r^2 h$
- Cone: Volume $= \dfrac{1}{3}\pi r^2 h$
- Sphere: Volume $= \dfrac{4}{3}\pi r^3$; Surface area $= 4\pi r^2$
- Pyramid: Volume $= \dfrac{1}{3} \times \text{base area} \times h$
Worked Example 4.1 — Surface Area of a Cylinder
A tin can has radius 4 cm and height 10 cm. Find its total surface area to the nearest square centimetre.
Step 1 Identify the components: two circular ends and a lateral (side) surface.
Step 2 Apply the formula: $$SA = 2\pi r^2 + 2\pi r h = 2\pi(4)^2 + 2\pi(4)(10)$$ $$= 32\pi + 80\pi = 112\pi \approx 352 \text{ cm}^2$$
The Pythagorean Theorem
Pythagorean Theorem
In a right triangle with legs $a$ and $b$ and hypotenuse $c$: $$a^2 + b^2 = c^2$$
This theorem is used to find any unknown side when the other two sides are known, and to determine whether a triangle is right-angled.
Worked Example 4.2 — Applying the Pythagorean Theorem
A ladder 5 m long leans against a wall. The base of the ladder is 2 m from the wall. How high up the wall does the ladder reach?
Step 1 Let $h$ be the height. Then $2^2 + h^2 = 5^2$.
Step 2 $4 + h^2 = 25 \Rightarrow h^2 = 21 \Rightarrow h = \sqrt{21} \approx 4.58$ m.
5. Financial Literacy
Ontario's MTH1W curriculum explicitly includes financial literacy. You will apply your algebra skills to real-world money problems involving interest, budgeting, and earning.
Simple Interest
Simple Interest Formula
$$I = Prt$$
where $I$ is the interest earned, $P$ is the principal (initial amount), $r$ is the annual interest rate (as a decimal), and $t$ is the time in years.
The total amount after $t$ years is: $A = P + I = P(1 + rt)$.
Worked Example 5.1 — Simple Interest
Mia invests $\$2{,}400$ at a simple interest rate of 3.5% per year. How much interest does she earn over 30 months?
Step 1 Convert time: 30 months $= \dfrac{30}{12} = 2.5$ years.
Step 2 Apply the formula: $I = Prt = 2400 \times 0.035 \times 2.5 = \$210$.
Mia earns $\$210$ in interest. Her total balance is $\$2{,}610$.
Compound Interest
Compound Interest Formula
$$A = P\left(1 + \frac{r}{n}\right)^{nt}$$
where $n$ is the number of compounding periods per year (annually $n=1$, semi-annually $n=2$, quarterly $n=4$, monthly $n=12$).
Worked Example 5.2 — Compound Interest
Noah invests $\$1{,}500$ at 4% per year, compounded quarterly, for 3 years. What is the final amount?
Step 1 Identify: $P = 1500$, $r = 0.04$, $n = 4$, $t = 3$.
Step 2 Apply: $$A = 1500\left(1 + \frac{0.04}{4}\right)^{4 \times 3} = 1500(1.01)^{12}$$ $$= 1500 \times 1.12683 \approx \$1{,}690.23$$
6. Data and Probability
Collecting and Organizing Data
Data can be primary (collected directly by you) or secondary (obtained from an existing source). It can be qualitative (categorical) or quantitative (numerical, either discrete or continuous).
Measures of Central Tendency
Mean, Median, and Mode
- Mean: the arithmetic average, $\bar{x} = \dfrac{\sum x_i}{n}$
- Median: the middle value when data is ordered; average of two middle values if $n$ is even
- Mode: the value that appears most often (a data set may have no mode, one mode, or multiple modes)
Worked Example 6.1 — Measures of Central Tendency
The following are test scores out of 50 for a class: 38, 42, 35, 42, 47, 31, 42, 38, 50, 44. Find the mean, median, and mode.
Mean $\bar{x} = \dfrac{38+42+35+42+47+31+42+38+50+44}{10} = \dfrac{409}{10} = 40.9$
Median Order: 31, 35, 38, 38, 42, 42, 42, 44, 47, 50. The 5th and 6th values are both 42. Median $= 42$.
Mode 42 appears 3 times. Mode $= 42$.
Probability
Theoretical Probability
$$P(A) = \frac{\text{number of favourable outcomes}}{\text{total number of equally likely outcomes}}$$
Probabilities range from 0 (impossible) to 1 (certain). The complement rule states: $P(\text{not } A) = 1 - P(A)$.
Worked Example 6.2 — Probability
A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. What is the probability of drawing a blue or green marble?
Solution Total marbles $= 10$. Favourable outcomes (blue or green) $= 3 + 2 = 5$.
$$P(\text{blue or green}) = \frac{5}{10} = \frac{1}{2} = 0.5 = 50\%$$
Alternatively, $P(\text{not red}) = 1 - P(\text{red}) = 1 - \dfrac{5}{10} = \dfrac{1}{2}$.
7. Practice Problems
Problem 1 — Exponent Laws
Simplify: $\dfrac{(3a^2b)^3}{9a^4b^2}$
Show Solution
Expand the numerator: $(3a^2b)^3 = 27a^6b^3$.
$$\frac{27a^6b^3}{9a^4b^2} = 3a^{6-4}b^{3-2} = 3a^2b$$
Problem 2 — Linear Equation
Solve for $x$: $\dfrac{2x - 1}{3} = \dfrac{x + 4}{2}$
Show Solution
Multiply both sides by the LCD (6): $2(2x - 1) = 3(x + 4)$.
$4x - 2 = 3x + 12$, so $x = 14$.
Problem 3 — Slope and Equation of a Line
Find the equation of the line passing through $A(2, 5)$ and $B(-4, -1)$ in slope-intercept form.
Show Solution
Slope: $m = \dfrac{-1-5}{-4-2} = \dfrac{-6}{-6} = 1$.
Using point $A(2,5)$: $y - 5 = 1(x - 2) \Rightarrow y = x + 3$.
Problem 4 — Volume
A cone has a base radius of 6 cm and a height of 8 cm. Find its volume to the nearest cubic centimetre.
Show Solution
$$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(6)^2(8) = \frac{1}{3} \times 36\pi \times 8 = 96\pi \approx 302 \text{ cm}^3$$
Problem 5 — Financial Literacy
Zara borrows $\$3{,}000$ at a simple interest rate of 6% per year. How long will it take for the interest to reach $\$540$?
Show Solution
Use $I = Prt$: $540 = 3000 \times 0.06 \times t$.
$540 = 180t \Rightarrow t = 3$ years.
Problem 6 — Polynomial Multiplication
Expand and simplify $(3x - 2)(2x + 5)$.
Show Solution
$(3x)(2x) + (3x)(5) + (-2)(2x) + (-2)(5)$
$= 6x^2 + 15x - 4x - 10 = 6x^2 + 11x - 10$
Problem 7 — Pythagorean Theorem
Determine whether a triangle with sides 9 cm, 40 cm, and 41 cm is a right triangle.
Show Solution
Check whether $a^2 + b^2 = c^2$ for the longest side as hypotenuse:
$9^2 + 40^2 = 81 + 1600 = 1681 = 41^2$. Yes, it is a right triangle.
Problem 8 — Probability
Two fair dice are rolled. What is the probability that the sum equals 8?
Show Solution
Total outcomes: $6 \times 6 = 36$. Favourable (sum = 8): (2,6), (3,5), (4,4), (5,3), (6,2) — that is 5 outcomes.
$$P(\text{sum} = 8) = \frac{5}{36} \approx 13.9\%$$